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Application of Vectors
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Force The force due to gravity, π π , pulls straight down on the box.
Part of the force, π 1 , pulls the box down the ramp. The other part of the force, π 2 , pulls the box against the ramp. The relationship can be expressed as π π = π 1 + π 2 π π π 2 π π
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hill/ breaking load scenario
These scenarios are about an object being parked on some sort of incline. π π is force due to gravity. In these scenarios, it is just the weight of the object. π 1 is the force that needs to be cancelled to ensure the object doesnβt roll down the hill π 2 is your force perpendicular to your hill. d is degree of incline π 1 π 2 d π π
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When looking for the force to cancel out πΉ 1
π π β sin π = π 1 π 1 π 2 π π
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Example An SUV weighing 5800 pounds is parked on a street that has an incline of 9Β°. Find the force required to keep the SUV from rolling down the hill. Round the force to the nearest hundredth. π 1 π 2 π π
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When looking for force perpendicular to the hill/ πΉ 2
π π β cos π = π 2 π 1 π 2 π π
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Example An SUV weighing 5800 pounds is parked on a street that has an incline of 9Β°. Find the force of the SUV perpendicular to the hill. Round the force to the nearest hundredth. π 1 π 2 π π
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Work The work, W, done by a force ,F , moving an object from point A to point B is π=πβ π΄π΅ = π π΄π΅ cos π
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Example: A child pulls a sled along level ground by exerting a force of 30 pounds on a rope that makes an angle of 35Β° with the horizontal. How much work is done pulling the sled 200 feet? Round answer to the nearest hundredth.
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Homework Vector Application WS 1
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